Flip a pile of unfair coins and count the heads. Do it over and over, and the counts stack into a hill. A new theorem says something strict about the top of that hill: the wider the hill, the gentler its fall is allowed to begin – but never gentler than a hard limit of one quarter.
No math background needed. Deeper layers open up when you want them.
Everything on this page comes down to one histogram and two numbers you can read off it.
Each coin has its own chance of landing heads – 50%, 20%, 80%, anything. Flip them all at once and count the heads. That count is random: sometimes 2, sometimes 5.
Repeat forever and tally the counts. The tallies form a histogram: the hill.
The hill only ever climbs, tops out, and falls – no dents, no double peaks. The first descent is the first bar that is shorter than the one before it: the step off the peak.
Two numbers describe the situation: the spread V (how wide the hill is) and the sharpness δ (how decisively the fall steepens right at that first step – 0 means coasting, 1 means dropping off a cliff to nothing).
Whenever the spread is at least 1, the theorem guarantees V · δ ≥ ¼. A wide hill may begin its fall gently; a narrow hill must fall hard. There is a fixed exchange rate between the two, and it is one quarter.
One consequence: past the peak, the bars must die off at least geometrically fast – each step multiplies the height by roughly 1 − 1/(4V) at most.
Your score is V · δ – spread times sharpness. Push it as low as you can while keeping the spread V ≥ 1. The theorem proves you can never get below 0.25. The best score anyone knows how to approach is ⅓ ≈ 0.333. Between the two lies open mathematics: a score under ⅓ would be a genuine discovery.
Tap a coin to select it, then drag the slider to change its heads chance. The gold slice shows the bias.
Same theorem, three altitudes. Switch whenever you like – nothing below depends on reading the harder ones.
Take any collection of coins, each with its own heads chance, flip them all, and count heads. The histogram of that count is a hill: it climbs, tops out, and falls, with no dips or double peaks. That much has long been known.
Now measure two things. First, the spread: how much the head-count wobbles from trial to trial. Each coin contributes its own wobble – a fair coin contributes the most (¼), and a coin that is nearly always heads or nearly always tails contributes almost nothing. Add the contributions up and you get the spread, V.
Second, the sharpness of the first descent. Just past the peak, each bar is some fraction of the bar before it – the bars shrink. The sharpness δ measures how much that shrinking accelerates right at the top: δ near 0 means the fall begins as a gentle coast; δ near 1 means the hill drops off a cliff.
The theorem: as long as the spread is at least 1, spread times sharpness is at least one quarter. You can trade one for the other – a very wide hill is allowed a very gentle start to its fall – but the product never dips below ¼. Ever. For any coins, any biases, any number of them.
Why demand spread ≥ 1? Because a single heavily biased coin makes a two-bar “hill” that is both narrow and cliff-like in a degenerate way, and the product can be tiny. Some minimum amount of genuine randomness is needed before the rule kicks in.
And here is the part nobody knows: the true unbreakable floor is somewhere between ¼ and ⅓. The theorem proves ¼ always works. Ordinary fair-ish binomial hills get scores as close to ⅓ as you like. Where the exact boundary sits is an open question – the one the game upstairs lets you poke at.
Let W = B₁ + … + Bₙ be a sum of independent coin flips, Bᵢ ∼ Bernoulli(pᵢ), with probabilities f_k = P(W = k) and variance V = Σ pᵢ(1−pᵢ). This is called a Poisson–binomial distribution.
A classical fact: the sequence (f_k) is log-concave – f_k² ≥ f_{k−1} f_{k+1} for every k. Equivalently, the consecutive ratios q_k = f_k / f_{k−1} only ever decrease. That is exactly why the histogram is a single hill.
The first descent is D = min{k : f_k < f_{k−1}}, the first index where the bars turn downward (the peak sits at D − 1). The normalized Turán deficit there is
Log-concavity says δ_D ≥ 0 – the classic Turán-type inequality, named for the same pattern Pál Turán found in Legendre polynomials. The new theorem upgrades “≥ 0” to a quantitative statement scaled by the variance:
Two immediate consequences. First, the same bound holds for the plain ratio drop: V(1 − f_{D+1}/f_D) ≥ ¼. Second, since ratios only decrease, the bars past the peak decay at least geometrically:
That is a tail bound you can read off three bars and the variance – no knowledge of the individual pᵢ required.
The proof is a squeeze between two forces, finished off by exact computer algebra. Suppose, for contradiction, that the deficit δ at the first descent is small – smaller than 1/(4V). The strategy is to show the hill would then have to be simultaneously too heavy and not heavy enough.
Smallness spreads. Cubic inequalities of Hillion and Johnson yield a recurrence between neighbouring deficits, δ_{k±1}(1 − δ_k) ≤ δ_k. Iterating it: if δ is small at the descent, the deficits stay small for about 1/δ steps in both directions. Small deficits mean the bar ratios barely change – so the bars near the peak are all forced to be a sizeable fraction of the peak mass p. The proof extracts explicit lower bounds L_r and R_r on f_{c∓r}/p around the peak c = D − 1.
Heavy and wide means high variance. The pairwise identity V = ½ Σᵢⱼ fᵢ fⱼ (i−j)² converts those mass bounds into V ≥ p² A(δ), where A(δ) collects the bounded terms. Meanwhile a lattice maximal-mass bound of Bobkov, Marsiglietti, and Melbourne runs the other way: V ≥ (p⁻² − 1)/12. A low peak already forces a large variance, so at any given variance the peak has a floor: p ≥ (1 + 12V)^{−1/2}.
The squeeze. Feeding the peak-mass lower bound into the variance lower bound eliminates p entirely and reduces the whole theorem to a single inequality about one number:
Certificates, not simulations. Substituting H = 1/δ − 1 splits the problem in two. For H ≥ 16 a uniform symbolic argument works: after clean substitutions, every relevant polynomial has visibly positive coefficients. For the compact window 3 < H ≤ 16, the inequality is proved cell by cell – thirteen cells, each reduced to an exact integer polynomial whose Bernstein expansion has strictly positive rational coefficients: 275 positive numbers in total. Because Bernstein basis functions are nonnegative, positive coefficients certify positivity of the polynomial. Everything is exact rational arithmetic – no floating point anywhere in the proof – and the certificates are replayed by an independently written checker program. Parts of the argument are additionally formalized in the Lean proof assistant, conditionally on the Hillion–Johnson recurrence.
Reference: Brett Reynolds, “A variance-scaled Turán inequality at the first descent of a Poisson–binomial mass function” (2026).
A map of the argument – each station is one move, in order. (Detail lives in the “whole story” tab above.)
Suppose the sharpness δ at the first descent were smaller than 1/(4V). The plan: show this forces a contradiction.
A recurrence between neighbouring deficits, δ_{k±1}(1 − δ_k) ≤ δ_k, means a gentle cliff forces gentle slopes for about 1/δ bars on either side. All those bars must stay close to peak height.
Total probability is fixed at 1, so many near-peak-height bars force the peak itself to be low, and the identity V = ½ ΣΣ fᵢfⱼ(i−j)² turns “low, broad, heavy” into a lower bound on the variance in terms of the peak mass.
A known maximal-mass bound pushes the other way: on the integers, variance V forces peak mass at least (1 + 12V)^{−1/2}. The two forces squeeze everything down to one inequality in δ alone.
The remaining one-variable inequality is proved symbolically for most of the range, and on a compact window by thirteen exact polynomial certificates – 275 strictly positive rational Bernstein coefficients, computed and independently re-checked in exact arithmetic. Contradiction reached; theorem proved.
Call the true unbreakable floor c★ – the lowest that spread × sharpness can ever be driven (with spread ≥ 1). The theorem proves c★ ≥ ¼. The tuned binomial family proves c★ ≤ ⅓. That is everything currently known.
A verified configuration scoring below ⅓ would improve the known upper bound – new mathematics. (The game runs on floating point, so treat any surprise as a lead to check exactly, not a proof.)
Measure just three bars at the top of the hill plus the overall spread, and the theorem hands you a certified tail bound: every later bar shrinks at least like e^{−r/(4V)}. Local data licenses a global inference – with the exact exchange rate stated.
Independent yes/no events with unequal chances are everywhere: which of n components fail, which of n voters turn out, which of n packets drop. Whenever you count successes across unequal, independent chances, this hill – and this rule about its top – is the shape you’re looking at.
Inequalities of the shape x_k² ≥ x_{k−1}x_{k+1} are named for Pál Turán, who found them in Legendre polynomials in the 1940s. The novelty here is not the pattern but the calibration: the gap in the inequality, at the one index where the hill turns, measured against the variance – with a universal constant.